Solution to Problem 109 Normal Stress

Free body diagram of Joint A
 

 

For wire AB: By sine law (from the force polygon):
$\dfrac{T_{AB}}{\sin 40^{\circ}} = \dfrac{W}{\sin 80^{\circ}}$

$T_{AB} = 0.6527W$

$\sigma_{AB} \, A_{AB} = 0.6527W$

$30(0.4) = 0.6527W$

$W = 18.4 \, \text{kips}$
 

For wire AC:
$\dfrac{T_{AC}}{\sin 60^{\circ}} = \dfrac{W}{\sin 80^{\circ}}$

$T_{AC} = 0.8794W$

$T_{AC} = \sigma_{AC} \, A_{AC}$

$0.8794W = 30(0.5)$

$W = 17.1 \, \text{kips}$
 

For safe load W,
$W = 17.1 \, \text{ kips}$           answer